air handling unit design

[ama_singh]

How to go about designing an AHU??i have the following data with me
inlet air conditions-
WBT-29 deg C
DBT -40 Deg C
Fan capacity-1000m3/hr
Oulet air conditions
RH-60%
air temp-25 deg C
how to calculate the heat load on the unit?
what amount of water condensation will take place?
Kindly guide me !

[mohamed khamis]

How to go about designing an AHU??i have the following data with me
inlet air conditions-
WBT-29 deg C
DBT -40 Deg C
Fan capacity-1000m3/hr
Oulet air conditions
RH-60%
air temp-25 deg C
how to calculate the heat load on the unit?
what amount of water condensation will take place?
Kindly guide me !

Hi ama_singh

Welcome to this forum. firstly are you sure about this data and what is the application in which u use this AHU. I think from this data the intake air is all fresh and the outlet air conditions is the indoor air conditions not the exit. Anyhow if u r sure about this data the design is :

first, The AHU will include cooling coil (Dx coil) and reheater :

the cooling coil capacity = (1000/3600)*1.2*(94.6-49.8) = 15.0 kW

The outlet cooling coil is of 18°C and 90% relative humidity
the removal moisture = (1000/3600)*1.2*(21.1-12.5) = 2.867 grams

The heater capacity is = (1000/3600)*1.2* (56.0-49.8) = 2.1 kW

so u locate the inlet and outlet conditions on pschrometeric chart then to reasch from the inlet to outlet u need two procees "cooling and dehimdified then reheating" . so i think u will need a cooling coil with capcity of 15 kw to 18 kW with face area of 20 inch * 8 inch R134a at te = 7.4C. i wish it could help

Cheers:)

[ama_singh]

actually we are to replace the catalyst in the reactor so to provide good air conditions in the reactor for the person to work inside it we will use the AHU.
the air will enter from one manhole and exit from another manhole of the reactor.it won't be recirculated.So how do i go about finding the heat load on the refrigeration unit based on the conditions i told you earlier.

[avdheshastri]

hello !
i m new in HVAC
can anyone tell me the following.

how to calculate heat load?
or
how to determine TR required for AHU?

please mail me on e-mail deleted

thankx

[nike123]

Make same research and then came here again with concrete questions:
http://www.natresnet.org/conference/2005/presentations/Storer.pdf
http://tinyurl.com/4kmhxh

[ashish]

Hello,

Deatails of AHU are as below :
Model NO. :- CCTU-040
Drawing No. :- 6070-1
Sr.No. :- 006040657
CFM/CMH :-3235/5500
Fan Size :- RDH-315R Qty :-02 Nos
Motor Kw/HP/P :- 3.7/5/2
Motor Pulley :-139X2A
Fan Pulley :- 140X2A
Condition required is : 25 deg 45% RH
Pl note :-Size of the room is 5.6X12.6X3 mtr.
We keep tomato paste at 80-85 deg inside the room and allow it to cool to 30-35 deg cel.
Pl suggest the solution as in winter we are not facing the problem but in summer it becomes an issue.


Ashish Khare




Hi ama_singh

Welcome to this forum. firstly are you sure about this data and what is the application in which u use this AHU. I think from this data the intake air is all fresh and the outlet air conditions is the indoor air conditions not the exit. Anyhow if u r sure about this data the design is :

first, The AHU will include cooling coil (Dx coil) and reheater :

the cooling coil capacity = (1000/3600)*1.2*(94.6-49.8) = 15.0 kW

The outlet cooling coil is of 18°C and 90% relative humidity
the removal moisture = (1000/3600)*1.2*(21.1-12.5) = 2.867 grams

The heater capacity is = (1000/3600)*1.2* (56.0-49.8) = 2.1 kW

so u locate the inlet and outlet conditions on pschrometeric chart then to reasch from the inlet to outlet u need two procees "cooling and dehimdified then reheating" . so i think u will need a cooling coil with capcity of 15 kw to 18 kW with face area of 20 inch * 8 inch R134a at te = 7.4C. i wish it could help

Cheers:)

[ednalds]

Hi ama_singh

Welcome to this forum. firstly are you sure about this data and what is the application in which u use this AHU. I think from this data the intake air is all fresh and the outlet air conditions is the indoor air conditions not the exit. Anyhow if u r sure about this data the design is :

first, The AHU will include cooling coil (Dx coil) and reheater :

the cooling coil capacity = (1000/3600)*1.2*(94.6-49.8) = 15.0 kW

The outlet cooling coil is of 18°C and 90% relative humidity
the removal moisture = (1000/3600)*1.2*(21.1-12.5) = 2.867 grams

The heater capacity is = (1000/3600)*1.2* (56.0-49.8) = 2.1 kW

so u locate the inlet and outlet conditions on pschrometeric chart then to reasch from the inlet to outlet u need two procees "cooling and dehimdified then reheating" . so i think u will need a cooling coil with capcity of 15 kw to 18 kW with face area of 20 inch * 8 inch R134a at te = 7.4C. i wish it could help

Cheers:)

can you please share the reference of your formula?
Also, AHU does not include Dx coils. AHU cooling medium is usually chilled water and the Dx coil is for PU or Package unit.

______
bad

[lana]

Hi there,

For AHU or any equipment dealing with air you need to know pychrometric calculations. I suggest that you talk to an experience consultant otherwise learn pychrometric first.
There are lots of issues involved to explain here.

[Polau]

Can someone tell me what is a supply and return means on a reefer panel?
I have a reefer which shows an al22 which is "evaporator safety"
The other problem the signal light for supply is off but the signal light for return is on. Tell me what I should do.

[panginies]

hi to all,im new here
let me know abt sabroe screw comp running parameter settings

[bangnhan87]

hi all, I'm a new member.
I'm designing an AHU with capacity'AHU:
cooling coil: 358kW at condition:
Inler air: DBT: 27 Deg C; RH:60%
Fan capacity-35000m3/hr. Pressure: 350Pa
outler air: DBT: 9 Deg C; RH:97%
Fan capacity-35000m3/hr. Pressure: 350Pa.


heat Load of room is 350kW, caculated by Trace700.

It has a proplem. that's AHU'dimension is small. about 2.5Dx2.1Wx3H. so, i put the return air at position 2.1Wx3H. and that so,the fan'space is not enough only remain 1.4m while the fan's dimension is 1300mm , the distance from the fan to the cooling coil would be 100mm. I just wonder if it changes the wind flow going into the fan.
what do you think about problem?

[Roadrunning123]

Hi there ok is it for your house or a commercial office ?
Yes the correct way is to gather all the info but you know what? There is an easier way but I need the total floor area and total windows and also is the premises in a real hot or cold part of your country.
Get this info and I'll give you a rock solid size system.

Cheers
Ps the amount of condensate is irrelevant as this will depend on the humidity of the day and how much of it is inside

[Vemuri.Murli]

Hi Sir,
Can you Please give the formulas used in finding the coil capacity?I am having the formulae Ma*(h1-h2)*60/14000 TR which is not matching with you.i am new to HVAC,i just want to correct my self



Hi ama_singh

Welcome to this forum. firstly are you sure about this data and what is the application in which u use this AHU. I think from this data the intake air is all fresh and the outlet air conditions is the indoor air conditions not the exit. Anyhow if u r sure about this data the design is :

first, The AHU will include cooling coil (Dx coil) and reheater :

the cooling coil capacity = (1000/3600)*1.2*(94.6-49.8) = 15.0 kW

The outlet cooling coil is of 18°C and 90% relative humidity
the removal moisture = (1000/3600)*1.2*(21.1-12.5) = 2.867 grams

The heater capacity is = (1000/3600)*1.2* (56.0-49.8) = 2.1 kW

so u locate the inlet and outlet conditions on pschrometeric chart then to reasch from the inlet to outlet u need two procees "cooling and dehimdified then reheating" . so i think u will need a cooling coil with capcity of 15 kw to 18 kW with face area of 20 inch * 8 inch R134a at te = 7.4C. i wish it could help

Cheers:)